Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. If it takes approximately ten minutes to serve each customer, find the mean use at least four digits after the decimal if rounding...)
and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) (use at least four digits after the decimal if rounding...) What is the probability that the total service time will exceed 2.5 hours? (use at least four digits after the decimal if rounding...)

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ShubhamSrivastava ShubhamSrivastava · Answer 1 · 2022-12-11T09:52:43+01:00

The answer of this question is found to be 0.0012.

The mean total service time is 10 minutes times lambda (the average number to turn up during the hour)

Lambda is 7, since the average rate is 7 per hour and your time window is 1 hour.

mean service time = 7 × 10 = 70 minutes.

For a poisson process, lambda is both the mean and variance of the arrivals.

So the variance = 10² × lambda = 700, since we want the variance of the service time, rather than the variance of the arrival rate.

This is because:

σ(aX) = a × σ(X).

and σ = sqrt(variance)

σ(service time) = 10 σ(arrival time)

variance(service time) = (10 × σ(arrival time))²

variance(service time) = 100 × variance(arrival time) = 100 × lambda = 100 × 7

mean(servicetime) = 70

variance(servicetime) = 700

2.5 hour = 150 min

Z = [tex]\frac{(S - 70)}{\sqrt{700} }[/tex] follow N(0,1)

P(S > 150) = P( [tex]\frac{(S - 70)}{\sqrt{700} }[/tex]> [tex]\frac{(150 - 70)}{\sqrt{700} }[/tex])

=P(Z > -2.5512)

=0.0012

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